CMO 1996 SOLUTIONS QUESTION Solution If f (x) = x3 − x − = (x − α)(x − β)(x − γ) has roots α, β, γ standard results about roots of polynomials give α + β + γ = 0, αβ + αγ + βγ = −1, and αβγ = Then S= N 1+α 1+β 1+γ + + = 1−α 1−β 1−γ (1 − α)(1 − β)(1 − γ) where the numerator simplifies to N = − (α + β + γ) − (αβ + αγ + βγ) + 3αβγ = − (0) − (−1) + 3(1) = The denominator is f (1) = −1 so the required sum is −7 QUESTION Solution For any t, ≤ 4t2 < + 4t2 , so ≤ than 4t2 < Thus x, y and z must be non-negative and less + 4t2 Observe that if one of x y or z is 0, then x = y = z = If two of the variables are equal, say x = y, then the first equation becomes 4x2 = x + 4x2 This has the solution x = 0, which gives x = y = z = and x = 1 which gives x = y = z = 2 Finally, assume that x, y and z are non-zero and distinct Without loss of generality we may assume that either < x < y < z < or < x < z < y < The two proofs are similar, so we only the first case We will need the fact that f (t) = 4t2 is increasing on the interval (0, 1) + 4t2 To prove this, if < s < t < then 4s2 4t2 − + 4t2 + 4s2 4t2 − 4s2 = (1 + 4s2 )(1 + 4t2 ) > f (t) − f (s) = So < x < y < z ⇒ f (x) = y < f (y) = z < f (z) = x, a contradiction Hence x = y = z = and x = y = z = are the only real solutions Solution Notice that x, y and z are non-negative Adding the three equations gives x+y+z = 4z 4x2 4y + + 2 + 4z + 4x + 4y This can be rearranged to give x(2x − 1)2 y(2y − 1)2 z(2z − 1)2 + + = + 4x2 + 4y + 4z Since each term is non-negative, each term must be 0, and hence each variable is either or The original equations then show that x = y = z = and x = y = z = are the only two solutions 2 Solution Notice that x, y, and z are non-negative Multiply both sides of the inequality y ≥0 + 4y by (2y − 1)2 , and rearrange to obtain y− 4y ≥ 0, + 4y and hence that y ≥ z Similarly, z ≥ x, and x ≥ y Hence, x = y = z and, as in Solution 1, the two solutions follow Solution As for solution 1, note that x = y = z = is a solution and any other solution will have each of x, y and z positive + 4x2 √ ≥ · 4x2 = 2x 4x2 and hence x ≥ = y, with equality if and only if = 4x2 – that is, x = Similarly, y ≥ z + 4x 1 with equality if and only if y = and z ≥ x with equality if and only if z = Adding x ≥ y, y ≥ z 2 and z ≥ x gives x+y +x ≥ x+y +z Thus equality must occur in each inequality, so x = y = z = The arithmetic-geometric mean inequality (or direct computation) shows that QUESTION Solution Let a1 , a2 , , an be a permutation of 1, 2, , n with properties (i) and (ii) A crucial observation, needed in Case II (b) is the following: If ak and ak+1 are consecutive integers (i.e ak+1 = ak ± 1), then the terms to the right of ak+1 (also to the left of ak ) are either all less than both ak and ak+1 or all greater than both ak and ak+1 Since a1 = 1, by (ii) a2 is either or CASE I: Suppose a2 = Then a3 , a4 , , an is a permutation of 3, 4, , n Thus a2 , a3 , , an is a permutation of 2, 3, , n with a2 = and property (ii) Clearly there are f (n − 1) such permutations CASE II: Suppose a2 = (a) Suppose a3 = Then a4 , a5 , , an is a permutation of 4, 5, , n with a4 = and property (ii) There are f (n − 3) such permutations (b) Suppose a3 ≥ If ak+1 is the first even number in the permutation then, because of (ii), a1 , a2 , , ak must be 1, 3, 5, , 2k − (in that order) Then ak+1 is either 2k or 2k − 2, so that ak and ak+1 are consecutive integers Applying the crucial observation made above, we deduce that ak+2 , , an are all either greater than or smaller than ak and ak+1 But must be to the right of ak+1 Hence ak+2 , , an are the even integers less than ak+1 The only possibility then, is 1, 3, 5, , ak−1 , ak , , 6, 4, Cases I and II show that f (n) = f (n − 1) + f (n − 3) + 1, n ≥ (∗) Calculating the first few values of f (n) directly gives f (1) = 1, f (2) = 1, f (3) = 2, f (4) = 4, f (5) = Calculating a few more f (n)’s using (*) and mod arithmetic, f (1) = 1, f (2) = 1, f (3) = 2, f (4) = 1, f (5) = 0, f (6) = 0, f (7) = 2, f (8) = 0, f (9) = 1, f (10) = 1, f (11) = Since f (1) = f (9), f (2) = f (10) and f (3) = f (11) mod 3, (*) shows that f (a) = f (a mod 8), mod 3, a ≥ Hence f (1996) ≡ f (4) ≡ (mod 3) so does not divide f (1996) QUESTION Solution Let BE = BD with E on BC, so that AD = EC: A D 4x 2x x x B 4x 2x C E By a standard theorem, CED and AD AB = ; CB DC so in CAB we have a common angle and CE AD AB CA = = = CD CD CB CB Thus CED ∼ CAB , so that ∠ CDE = ∠ DCE = ∠ ABC = 2x Hence ∠BDE = ∠ BED = 4x , whence 9x = 180◦ so x = 20◦ Thus ∠A = 180◦ − 4x = 100◦ Solution Apply the law of sines to ABD and AD sin x = BD sin 4x BDC to get and 1+ AD BC sin 3x = = BD BD sin 2x Now massage the resulting trigonometric equation with standard identities to get sin 2x (sin 4x + sin x) = sin 2x (sin 5x + sin x) Since < 2x < 90◦ , we get 5x − 90◦ = 90◦ − 4x , so that ∠A = 100◦ QUESTION Solution Let m f (n) = n − [rk n] k=1 m k=1 m = m rk − =n [rk n] k=1 {rk n − [rk n]} k=1 Now ≤ x − [x] < 1, and if c is an integer, (c + x) − [c + x] = x − [x] Hence ≤ f (n) < m = m Because f (n) is an integer, ≤ f (n) ≤ m − k=1 To show that f (n) can achieve these bounds for n > 0, we assume that rk = integers; ak < bk Then, if n = b1 b2 bm , (rk n) − [rk n] = 0, k = 1, 2, , m and thus f (n) = Letting n = b1 b2 bn − 1, then rk n = rk (b1 b2 bm − 1) = rk {(b1 b2 bm − bk ) + bk − 1)} = integer + rk (bk − 1) This gives rk n − [rk n] = rk (bk − 1) − [rk (bk − 1)] = ak ak (bk − 1) − (bk − 1) bk bk = ak − ak bk − ak − = ak − ak bk − (ak − 1) =1− Hence f (n) = ak bk ak = − rk bk m k=1 (1 − rk ) = m − ak where ak , bk are bk